Can we solve for $g$ when $\varepsilon$ is small?
$\newcommand{\sinc}{\operatorname{sinc}}$ $$3\sinc\left(-1+ \frac\varepsilon T \right)-3\sinc\left(1+\frac\varepsilon T\right)-\sinc\left(-3+\frac\varepsilon T\right)+\sinc\left(3+\frac\varepsilon T\right) = \frac\varepsilon g$$
I am not able to get a constant value for $g$. Can one do this?
Because $\operatorname{sinc}$ is an even function, for $\epsilon=0$ the left side is zero. Is $\epsilon$ supposed to be small? If so, we can use $\dfrac d{dx} \operatorname{sinc}(x)=\dfrac {x \cos x-\sin x}{x^2}$ so you can evaluate the first-order Taylor expansion. The first term will contribute $3\dfrac \epsilon T (- \cos 1+\sin 1)$ as will the second. You get $6\dfrac \epsilon T (- \cos 1+\sin 1)+2\dfrac \epsilon {9T} (- 3\cos 3+\sin 3)=\dfrac \epsilon g$ The epsilons divide out and you can invert both sides to find $g$